Question

The student's lab manual says to mix 50 mL of his 2.0 M CaCl2 solution with 50 mL of a 3.0 M aqueous solution of sodium carbonate(Na2CO3)
a. Write a balanced equation to show the reaction. Include state symbols
b. What kind of reaction is this?
c. What is the limiting reagent?
d. How many grams of precipitate should he get?
e. If he got 8.29 of product, what is his percent yield?

Answer 1

CaCl2+Na2CO3-->CaCO3+2NaCl
A double displacement reaction is a type of reaction where part of of one reactant is replaced by part of another reactant.
The calcium traded its chloride ion for the sodium's carbonate ion.

Answer 2

what are the state symbols though? (s) and (aq)?

Answer 3

CaCl2(aq)+Na2CO3(aq)-->CaCO3(s)+2NaCl(aq)

Answer 4

i think you are right calcium carbonate is insoluble.

Answer 5

and what is a limiting regent..?

Answer 6

The reactant in a chemical reaction that limits the amount of product that can be formed.
Convert Volume units from cm^3 to dm^3
1cm^3=0.001 dm^3
Use Number of moles=Concentration(mol/dm^3)*Volume(dm^3)
Number of moles of CaCl2=(50/1000)*2=0.1
Number of moles of Na2CO3=(50/1000)*3=0.15
Those Values represent the actual number of moles present from each reactant. Now find the the number of moles of CaCl2 based on Na2CO3 acutal no. of moles. this is also done to Na2CO3 based on CaCl2 actual no. of moles present.
We do that using mole ratio based on the balanced chemical equation, if you find the theoretical number of moles of reactant is greater than the actual number of moles this reactant will be the limiting reagent.

Answer 7

a. CaCl2 + Na2CO3 → CaCO3(s) + 2 NaCl
b. It's a double displacement reaction with a precipitate.
c. (50 mL) x (2.0 M CaCl2) = 100 mmol CaCl2
(50 mL) x (3.0 M NaHCO3) = 150 mmol Na2CO3
100 millimoles of CaCl2 would react completely with 100 x (1/1) = 100 millimoles of Na2CO3, but there is more Na2CO3 present than that, so Na2CO3 is in excess and CaCl2 is the limiting reagent.
d. (100 mmol CaCl2) x (1 mol CaCO3 / 1 mol CaCl2) x (100.0875 g CaCO3/mol) =10008.75 mg = 10 g CaCO3
e. Supposing the missing units on "8.29" to be "grams":
(8.29 g) / (10 g) = 0.829 = 83% yield